Proposition 5.37.14.B929E98153F2 Let $\psi $ and $\phi $ be plurisubharmonic functions on $\Omega $ with $\phi <0$, and let $p>1$. For holomorphic $f$ with $I=\| f\| _{\psi +\phi }^2<+\infty $, define
Then $c\leq (1-1/p)I$.
B929E98153F2Proposition 5.37.14.B929E98153F2 Let $\psi $ and $\phi $ be plurisubharmonic functions on $\Omega $ with $\phi <0$, and let $p>1$. For holomorphic $f$ with $I=\| f\| _{\psi +\phi }^2<+\infty $, define
Then $c\leq (1-1/p)I$.
Proof. For $t>0$, apply lemma 5.37.10 with
Their exponential sum is $S=(1+t\mathrm e^{p\phi })/(1+t)<1$. The integral $J$ of the lemma is finite, since $\| f\| _\psi ^2\leq I$ and $S\geq 1/(1+t)$. The first component $F_1$ belongs to $\Gamma (\Omega ,\mathcal I(p\phi +\psi ))$, so the second component $F_2=f-F_1$ is admissible for $c$, and the norm estimate gives
Set $t=x^ p$ and integrate over $x>0$. With $\kappa _ k=\int _0^\infty (1+x^ p)^{-k}\, \mathrm{d}x$, Tonelli's theorem and the change of variable $x\mapsto x\mathrm e^\phi $ give $c\kappa _1\leq I\kappa _2$; the pole set of $\phi $ is Lebesgue null. Since $p>1$, both integrals are finite, and integrating the derivative of $x/(1+x^ p)$ over $(0,\infty )$ gives $\kappa _2=(1-1/p)\kappa _1$. The argument does not require the infimum to be attained. $\square$
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